What volume of 0.25 M HNO₃ is necessary to react exactly with 7.4 g of Ca(OH)₂? A) 8.0 × 10² mL B) 1.2 × 10² mL C) 4.0 × 10² mL D) 2.5 × 10² mL E) 2.0 ×10² mL

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First, we need to correct the chemical formulas: HNO3 for nitric acid and Ca(OH)2 for calcium hydroxide. The balanced chemical equation for the reaction between HNO3 and Ca(OH)2 is:$$ \text{Ca(OH)}_2 + 2\text{HNO}_3 \rightarrow \text{Ca(NO}_3)_2 + 2\text{H}_2\text{O} $$From the equation, 1 mole of Ca(OH)2 reacts with 2 moles of HNO3. The molar mass of Ca(OH)2 is approximately 74 g/mol, so 7.4 g corresponds to 0.1 mol. Therefore, 0.2 mol of HNO3 is needed.Given the concentration of HNO3 is 0.25 M, the volume $V$ needed can be calculated using the formula $n = C \times V$, where $n$ is the amount of substance in moles, $C$ is the concentration, and $V$ is the volume in liters. Rearranging for $V$, we get $V = \frac{n}{C}$.$$V = \frac{0.2 \, \text{mol}}{0.25 \, \text{M}} = 0.8 \, \text{L} = 800 \, \text{mL}$$Therefore, 800 mL (or 8.0 × 10^2 mL) of 0.25 M HNO3 is necessary, which corresponds to choice A.

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